TITLE: Five-line inscription: Equilateral triangle, one of whose points is given, in circle with unknown center
QUESTION [7 upvotes]: PROBLEM Here is a surprisingly intriguing challenge posed on Euclidea, a mobile app for Euclidean constructions. I have a geometric construction that achieves the goal but not within the constraints. Could anyone advise on a construction that would fit within these constraints? 

Given A point $P$ on a circle $O$ whose center is not given.
Goal Inscribe an equilateral triangle $\Delta{PQR}$ in $\bigcirc{O}$.
Constraints Construct all three sides of the $\Delta{PQR}$ to achieve the goal. Only an unmarked straightedge and a non-rusty collapsible compass (i.e., a compass that, while on paper, can have any radius but which cannot maintain said radius when not on paper) can be used. Use only from the construction steps below a number of steps that together require just five lines in total.

Construct a point: 0 lines (L).
Mark the intersection of two curves with a point: 0L.
Construct a new line (*or line segment or ray): 1L.
Extend a given line segment (*or ray): 1L.
Construct a circle (non-rusty collapsible compass): 1L.
Construct the perpendicular bisector of a line segment: 1L.
Construct a new line perpendicular to an old line: 1L.
Construct an angle bisector: 1L.

Hints from Euclidea A known solution employs five constructions in the order: Circle, circle, perpendicular bisector, perpendicular bisector, and line.

PREVIOUS WORK As mentioned above, I have a construction that matches neither the 5L-constraint nor the hint. The following figure accompanies the six-line construction below, where the given and goal are shown as an inset:


Construct a point $A \ne P$ on circle $\bigcirc{O}$ [0L running total].
Construct $\bigcirc{A(P)}$ centered on point $A$ and with radius $AP$ [1L running total].


$\bigcirc{A(P)}$ also intersects $\bigcirc{O}$ at another point $B \ne P$.

Construct $\bigcirc{P(B)}$ [2L running total].


$\bigcirc{P(B)}$ also intersects $\bigcirc{A(P)}$ at another point $C \ne B$.
Therefore, $\bigcirc{P(C)} = \bigcirc{P(B)}$.
Also, $\overline{CP}$ is tangent to $\bigcirc{O}$ at point $P \Longrightarrow m\angle{CPO} = 90^\circ{}$.

Construct $\bigcirc{C(P)}$ [3L running total].


$\bigcirc{C(P)} \cap \bigcirc{P(C)} = \lbrace D, E\rbrace$.
$\bigcirc{C(P)} \cong \bigcirc{P(C)} \Longrightarrow m\angle{CPD} = m\angle{CPE} = 60^\circ{} \Longrightarrow m\angle{DPE} = 120^\circ{}$.
Without actually constructing anything: Let $\overrightarrow{EP} \cap \bigcirc{P(C)} = E' \Longrightarrow$ $m\angle{DPE'} = m\angle{EPE'} - m\angle{DPE} = 180^\circ{} - 120^\circ{} = 60^\circ{}$.
Therefore, $\angle{DPE'}$ is the first corner of the desired triangle.

Construct $\overline{DP}$ [4L running total].


$\overline{DP} \cap \bigcirc{O} \cap \neg P = Q$.
$\overline{PQ}$ is the first side of the desired triangle.

Construct $\overline{DP}$ [5L running total].


$\overline{EP} \cap \bigcirc{O} \cap \neg P = R$.
$\overline{PR}$ is the second side of the desired triangle.

Construct $\overline{QR}$ [6L running total].


$\overline{QR}$ is the last side of the desired triangle. $\blacksquare$

REPLY [2 votes]: Here is a proof of the construction by @Blue:
 

$\angle{B_1C_1P} = 60^\circ$ is bisected by $\overline{C_1C_2} \Longrightarrow \angle{C_2C_1P} = 30^\circ$.
Let $x := \angle{B_2C_1P} \Longrightarrow \angle{B_2C_1B_2} = \angle{B_1C_1P} + \angle{B_2C_1P} = 30^\circ + x$.
$\overline{B_2C_1} \perp \overline{PQ}$ and $\mathbf{(2)} \Longrightarrow \angle{PST} = 60^\circ - x \Longrightarrow \angle{SPT} = 30^\circ + x$.
Inscribed Angle Theorem and $\mathbf{(2)} \Longrightarrow \angle{C_2PB_2} = 2\angle{B_2C_1C_2} = 60^\circ + 2x$.
$\overline{PR}$ is the perpendicular bisector of $\overline{B_2C_2}$ and $\mathbf{(4)} \Longrightarrow \angle{B_2PR} = 30^\circ + x$. 
$\overline{PR}$ is the perpendicular bisector of $\overline{B_2C_2}$ and $\mathbf{(4)} \Longrightarrow \angle{C_2PR} = 30^\circ + x$.
$\overline{PS} \perp \overline{B_2C_1} \Longrightarrow \angle{C_1SP} = 60^\circ - x$. 
From $\mathbf{(3)}$, $\angle{B_1PC_2} = \angle{C_2PS} + \angle{SPT} = \angle{C_2PS} + (30^\circ + x) = 60^\circ$ $\Longrightarrow \angle{C_2PS} = 30^\circ - x$.
By $\mathbf{(6)}$ and $\mathbf{(8)}$, $\angle{QPS} = 60^\circ$.
From $\mathbf{(3)}$ and $\mathbf{(5)}$, $\angle{B_1PO} = \angle{B_2PO} \Longrightarrow \angle{TPS} + \angle{OPQ} = \angle{B_2PR} + \angle{OPR}$  $\Longrightarrow \angle{OPQ} = \angle{OPR}$.
SSA congruence, $\mathbf{(10)}$, $OP = OP$, $OQ = OR \Longrightarrow \Delta{OPQ} \cong \Delta{OPR} \Longrightarrow PQ = PR$.
From $\mathbf{(9)}$ and $\mathbf{(11)}$, $\Delta{PQR}$ is equilateral. QED. $\blacksquare$